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melomori [17]
3 years ago
13

Why is measuring the duration of a number of swings a better way to determine the period of a pendulum than by measuring a singl

e cycle?
Physics
1 answer:
Alekssandra [29.7K]3 years ago
5 0

Answer:

It is to reduce the expected relative error of the measurement.

Explanation:

If there was a way to measure without error, this method would be unnecessary. In practice, the pesky error is always there. The sources are varied: inexact instrument, small inaccuracies in starting/stopping the timer, etc. But, it is reasonable to assume that such an error is random and has an expected spread that is <em>independent</em> of the actual duration of measurement. Under such assumptions, the methods offers a great advantage:

Let ε denote an additive measurement error. Let the error be random, symmetric (negative/positive), distributed in some fixed range independent of the actual measured value. The error represents an additive component in our measurement, i.e., (measurement) = (true value) + (error). In the case of one period T, we get to measure the duration T':

T' = T + \epsilon

so the relative error is

\frac{|T'-T|}{T}=\frac{|T+\epsilon-T|}{T}=\frac{|\epsilon|}{T}

In a separate experiment, suppose you measure n periods. Same error applies:

T_n'=n\cdot T+\epsilon

we can get a single period by dividing the measured value by n:

\frac{T_n'}{n}=\frac{n\cdot T +\epsilon}{n}=T+\frac{\epsilon}{n}

and the relative error of such a result will be:

\frac{|T+\frac{\epsilon}{n}-T|}{T}=\frac{|\epsilon|}{n\cdot T}

which is n times smaller than the relative error of the single measurement above. The more periods are included in the measurement, the smaller the expected error!

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A projectile is fired upward with an initial speed vo on an airless world. A short time later, it comes back down and has a fina
Zinaida [17]

Answer:

W_{grav} < 0

Explanation:

When a projectile is fired upwards with some initial speed then the it reaches the top of the projectile and then falls back to the ground.

According to the question we need to find the work done by the gravity which is acting downwards for the projectile when it is at a position just about to hit the ground in course of falling down.

As we know that work is given as:

W=F.s\cos\theta

here:

F= force of gravity on the object (which is acting downwards)

s= displacement of the object

  • Here the work done by the gravity at an instant just before the projectile hits the earth will be negative as the displacement is in the direction opposite to the force of gravity.
7 0
4 years ago
Why am i getting so many spam calls on my cell phone.
OleMash [197]

Answer:

Why do I keep getting spam calls? Experts credit the ascendance of spam phone calls to fundamental problems with caller ID, a phone system where anyone can operate as a carrier, the inability to detect bad callers, and a number of bad actors exploiting those flaws to drive billions of calls to American phones.

Explanation:

5 0
2 years ago
Why is a microscope or magnifying glass used to view objects close up instead of moving the object closer to your
Shalnov [3]

Answer:

D; The microscope and magnifying glass block out the light, which allows the naked eye to focus on the object.

Explanation:

This is to prevent chromatic abberation

8 0
3 years ago
The resistanceless inductor is connected across the ac source whose voltage amplitude is 24.5 V and angular frequency is 850 rad
timurjin [86]

Answer:

Part A 2.88 A, PART B 28.82 mA PART C  0.288 mA

Explanation:

We have given angular frequency \omega =850rad/sec

Voltage source has an amplitude of 24.5 V, So V= 24.5 volt

Part A

We have given inductance L=10^{-2}H

So inductive reactance X_L=\omega L=850\times 10^{-2}=8.5ohm

So current i=\frac{V}{X_L}=\frac{24.5}{8.5}=2.8823A

Part B

We have given inductance L=1 H

So inductive reactance X_L=\omega L=850\times 1=850ohm

So current i=\frac{V}{X_L}=\frac{24.5}{850}=28.82mA

Part C

We have given inductance L=100 H

So inductive reactance X_L=\omega L=850\times 100=85000ohm

So current i=\frac{V}{X_L}=\frac{24.5}{85000}=0.288mA

8 0
3 years ago
To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T
Fantom [35]

Answer:

V_1= 3.4*10^7m/s

Explanation:

From the question we are told that

Nucleus diameter d=5.50-fm

a 12C nucleus

Required kinetic energy K=2.30 MeV

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

            K_2 +U_2=K_1+U_1

where

K_1 =initial kinetic energy

K_2 =final kinetic energy

U_1 =initial electric potential

U_2 =final electric potential

mathematically

   U_2 = \frac{Kq_pq_c}{r_2}

where

r_f=distance b/w charges

q_c=nucleus charge =6(1.6*10^-^1^9C)

K=constant

q_p=proton charge

Generally kinetic energy is know as

         K=\frac{1}{2}  mv^2

Therefore

         U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2}  mv_1^2 +U_1

Generally equation for radius is d/2

Mathematically solving for radius of nucleus

         R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})

         R=2.75*10^-^1^5m

Generally we can easily solving mathematically substitute into v_1

   q_p=6(1.6*10^-^1^9C)

   K_1=9.0*10^9 N-m^2/C^2

   U_1= 0

   R=2.75*10^-^1^5m

   K=2.30 MeV

   m= 1.67*10^-^2^7kg

   V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }

    V_1= 3.4*10^7m/s

Therefore the proton must be fired out with a speed of V_1= 3.4*10^7m/s

8 0
3 years ago
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